Find a non-normal operator that is similar to a normal operator

Question

Find a non-normal operator that is similar to a normal operator

My Solution:

For non-normal operator consider, $$T=\begin{bmatrix} 1&1\\0&1\end{bmatrix} .$$ and For normal operator consider, $$I=\begin{bmatrix} 1&0\\0&1\end{bmatrix} .$$

Can we say that: they are similar?

Answer 1

Let we take, $$A=\begin{bmatrix} 2&3\\0&5\end{bmatrix} $$ and $$B=\begin{bmatrix} 2&0\\0&5\end{bmatrix} $$ Here, $B$ is self-adjoint matrix, so it is normal also! And $A$ is non normal matrix. Clearly, $A$ and $B$ are similar.

Answer 2

No, these matrices are not similar. In particular, the only matrix that is similar to the identity matrix is the identity matrix itself: for any invertible matrix $S$, $SIS^{-1} = SS^{-1} = I$.

Hint: You are on the right track. An upper triangular matrix will be normal if and only if it is diagonal. On the other hand, a matrix will be similar to a normal matirix if and only if it is diagonalizable.